Written out explicitly, cubic code is the translation-invariant negative sum of two types of interaction terms as in Figure5.1. There are two qubits per site, and the two-letter notation stands for tensor product of Pauli matrices. For example,XI =σx⊗I,ZZ =σz⊗σz, etc. The third cube specifies the coordinate system of the simple cubic lattice. The generating map for the stabilizer module is
σcubic-code=
1 +x+y+z 0
1 +xy+yz+zx 0
0 1 + ¯x¯y+ ¯yz¯+ ¯z¯x 0 1 + ¯x+ ¯y+ ¯z
IZ
ZI
ZI ZZ
II IZ
IZ ZI
IX
XI
XI II
XX IX
IX XI
z
yz
xz xyz
1 y
x xy
Figure 5.1: Stabilizer generators of the 3D cubic code. HereX ≡σx andZ ≡σx represent single- qubit Pauli operators, while I is the identity operator. Double-letter indices represent two-qubit Pauli operators, for example,IZ≡I⊗Z,ZZ ≡Z⊗Z,II ≡I⊗I, etc.
where one has to interchange the first and second qubit. The associated ideal is contained in a prime ideal of codimension 2 inF2[x±1, y±1, z±1]:
I(σ)⊆(1 +x+y+z, 1 +xy+yz+zx) =pxyz.
Since codimI(σ)≥2, the characteristic dimension is 1. Since cokercubic-code=R/pxyz⊕R/pxyz, any nonzero element ofpxyz is a fractal generator. Since the conditions used in the search for the model were only necessary conditions for the no-strings rule. A rigorous treatments is as follows.
We prove a purely algebraic statement, of which the no-strings rule is an interpretation. A more elementary method can be found in [50].
Lemma 5.3.1. LetS=F2[x, y, z]be a polynomial ring, andp= (1+x+y+z, 1+xy+yz+zx)⊆San ideal. Ifm1e1+m2e2∈p for polynomialse1, e2 and monomialsm1, m2 such thatgcd(m1, m2) = 1, then only one of the following is true:
• e1∈p ande2∈p.
• max(dege1,dege2)≥max(degm1,degm2).
Proof. Put F4 = {0,1, ω, ω2}, i.e., ω is the primitive third root of unity over the binary field.
Consider a ring homomorphismφ:S→U :=F4[t] defined by
φ:x7→1 +t, y7→1 +ωt, z7→1 +ω2t.
It maps the two generators ofpto zero inU.
φ(1 +x+y+z) = 4 + (1 +ω+ω2)t= 0,
φ(1 +xy+yz+zx) = 4 + 2(1 +ω+ω2)t+ (1 +ω+ω2)t2= 0.
Moreover, kerφis preciselyp. (This can be verified by eliminating the variablexand computing the Gr¨obner basis of ((y+1)+ω(z+1), ω2+ω+1) in an elimination monomial order.) Ifm1e1+m2e2∈p, thenφ(m1)φ(e1) =φ(m2)φ(e2). Sincem1 andm2 are co-prime monomials andφ(x), φ(y), φ(z) are pairwise co-prime, it follows thatφ(m1) andφ(m2) are nonzero and co-prime. Therefore,φ(e1) = 0 if and only if φ(e2) = 0 if and only ife1, e2 ∈ p, which is the first case. If φ(e1)6= 0, then φ(m1) must divide φ(e2) and φ(m2) must divide φ(e1). Since φ is degree-preserving, we have the second case.
Theorem 5.1. The cubic code obeys the no-strings rule with the constantα= 1under `∞-metric.
Proof. Since the cubic code is translationally invariant, we may use the formalism of Chapter 3.
Since the cubic code is of CSS type, where cokeris a direct sum of isomorphic summands, we only
have to considerσx-type charges. The set of all virtual charges isR1=R=F2[x±1, y±1, z±1] and the set of all trivial charges are given by a submodule (ideal)pxyz= (1 +x+y+z,1 +xy+yz+zx)⊆R.
Note thatRis the localization ofS=F2[x, y, z] by a single elementxyz, andpxyzis the localization ofp= (1 +x+y+z, 1 +xy+yz+zx) by the same single elementxyz.
Lete1 ande2 be the charges contained in two boxes of a string segment. Since they are overall trivial, we have e1+xiyjzke2 ∈ p wherei, j, k ∈Z. Equivalently, we may writem1e1+m2e2 ∈ p wheree1ande2have nonnegative exponents, andm1, m2are monomials such that each variable (x, y, orz) appears only in one ofm1 or m2. Sincepxyz∩S=p, which can be verified using Gr¨obner basis techniques, we are in the situation of Lemma 5.3.1. The width of the string segment is the maximum of the degrees ofe1and e2, and the`1-distance between the boxes enclosinge1ande2 is
≤max(degm1,degm2). If we use`∞-distance, the constant αin the no-strings rule is 1.
Let us explicitly calculate the ground-state degeneracy when the Hamiltonian is defined on L×L×Lcubic lattice with periodic boundary conditions. By Corollary3.4.5,
k= dimF2R/(pxyz+bL)⊕R/(pxyz+bL) = 2 dimF2R/(pxyz+bL),
wherebL= (xL−1, yL−1, zL−1). So the calculation of ground-state degeneracy comes down to the calculation of
d= dimF2T0/p whereT0=F2[x, y, z]/(xn1−1, yn2−1, zn3−1).
We may extend the scalar field to any extension field without changingd. LetFbe the algebraic closure ofF2 and let
T=F[x, y, z]/(xn1−1, yn2−1, zn3−1)
be an Artinian ring. By Proposition3.4.3, it suffices to calculate for each maximal idealmofT the vector space dimension
dm = dimF(T /p)m
of the localized rings, and sum them up.
Supposen1, n2, n3>1. By Nullstellensatz, any maximal ideal ofT is of form m= (x−x0, y− y0, z−z0) wherexn01 =yn02 =z0n3 = 1. (Ifn1=n2=n3= 1, thenT becomes a field, and there is no maximal ideal other than zero.) Putni= 2lin0iwheren0i is not divisible by 2. Since the polynomial xn1−1 contains the factorx−x0with multiplicity 2l1, it follows that
Tm =F[x, y, z]m/(x2l1+a0, y2l2 +b0, z2l3 +c0)
wherea0=x20l1, b0=y02l2, c0=z20l3. Hence, (T /p)m∼=F[x, y, z]/I0 where
I0 = (x+y+z+ 1, xy+xz+yz+ 1, x2l1 +a0, y2l2 +b0, z2l3 +c0).
IfI0=F[x, y, z], then dm = 0.
Without loss of generality, we assume that l1 ≤l2 ≤l3. By powering the first two generators of I0, we see that (x0, y0, z0) must be a solution of them in order for I0 not to be a unit ideal.
Eliminatingz and shiftingx→x+ 1,y→y+ 1, our objective is to calculate the Gr¨obner basis for the proper ideal
I= (x2+xy+y2, x2l1 +a, y2l2+b) wherea=a0+ 1 and b=b0+ 1. So
dm= dimFF[x, y]/I.
One can easily deduce by induction that y2m +x2m−1(mx+y) ∈I for any integer m ≥ 0. And b=ωa2l2−l1 for a primitive third root of unityω. So we arrive at
I= (y2+yx+x2, yx2l2−1+b(1 +l2ω2), x2l1 +a)
We apply the Buchberger criterion. Ifa6= 0, i.e.,x06= 1, thenb6= 0 andI= (x+(ω2+l2)y, x2l1+a), sodm= 2l1
If a=b= 0, thenI= (y2+yx+x2, yx2l2−1, x2l1). The three generators form Gr¨obner basis if l2=l1. Thus, in this case,dm= 2l1+1−1. If l2> l1, thendm= 2l1+1.
To summarize, except for the special point (1,1,1) ∈ F3 of the affine space, each point in the algebraic set
V =
(x, y, z)∈F3
x+y+z+ 1 =xy+xz+yz+ 1 = 0 xn01−1 =yn02−1 =zn03−1 = 0
contribute 2l1 tod. The contribution of (1,1,1) is either 2l1+1or 2l1+1−1. The latter occurs if and only ifl1 and l2, the two smallest numbers of factors of 2 inn1, n2, n3, are equal. Let d0 = #V be the number of points inV. The desired answer is
d= 2l1(d0−1) +
2l1+1−1 ifl1=l2
2l1+1 otherwise wherel1≤l2≤l3 are the number of factors of 2 inni.
The algebraic set defined by (x+y+z+ 1, xy+xz+yz+ 1) is the union of two isomorphic
lines intersecting only atx=y=z= 1, one of which is parametrized byx∈Fas
(1 +x,1 +ωx,1 +ω2x)∈F3, and another is parametrized as
(1 +x,1 +ω2x,1 +ωx)∈F3.
whereωis a primitive third root of unity. Therefore, the purely geometric numberd0= 2d1−1 can be calculated by
d1= degxgcd
(1 +x)n01+ 1,(1 +ωx)n02+ 1,(1 +ω2x)n03+ 1 .
Using (α+β)2p = α2p+β2p and ω2+ω+ 1 = 0, one can easily compute some special cases as summarized in the following corollary. Some values ofk for smallLare presented in Table5.2and Figure5.2.
20 40 60 80 100 L
100 200 300 400 k
200 400 600 800 1000L
1000 2000 3000 4000 k
Figure 5.2: Number of encoded qubitskof the cubic code defined onL×L×Lperiodic lattice
Corollary 5.3.2. Let2k be the ground-state degeneracy of the cubic code on the cubic lattice of size L3 with periodic boundary conditions. (k=k(L)is the number of encoded qubits.) Then
k+ 2
4 = degxgcd (1 +x)L+ 1, (1 +ωx)L+ 1, (1 +ω2x)L+ 1
F4
=
1 if L= 2p+ 1, L if L= 2p, L−2 if L= 4p−1, 1 if L= 22p+1−1.
whereω2+ω+ 1 = 0andp≥1 is any integer. IfL= 2rL0, thenk(L) + 2 = 2r(k(L0) + 2).
(k+ 2)/4 = 1 + 12P
nqn(L) qn(L) is nonzero only ifn|L.
qn(L) n
1 15 = 24−1 = 3·5
5 63 = 26−1 = 32·7
20 255 = 28−1 = 3·5·17
80 1023 = 210−1 = 3·11·31
322 4095 = 212−1 = 32·5·7·13 5 341 = (210−1)/(22−1) = 11·31 6 1365 = (212−1)/(22−1) = 3·5·7·13 49 5461 = (214−1)/(22−1) = 43·127
4 455 = (212−1)/(23+ 1) = 5·7·13 3 585 = (212−1)/(23−1) = 32·5·13 9 9709 = (218−1)/(3(23+ 1)) = 7·19·73 5 11275 = 11(210+ 1) = 52·11·41
Table 5.2: Numerical values of k for odd linear size L computed from Corollary 5.3.2. The list is complete if 2≤L≤20000. For example, if L= 945 = 15·63, then k+24 = 1 + 12·(1 + 5) = 73.